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In a triangle ABC if cot(A/2)cot(B/2)=c,...

In a triangle `ABC` if `cot(A/2)cot(B/2)=c, cot(B/2)cot(C/2)=a ` and `cot(C/2)cot(A/2)=b` then `1/(s-a)+1/(s-b)+1/(s-c)=`

A

`-1`

B

0

C

1

D

2

Text Solution

Verified by Experts

The correct Answer is:
D
`cot.(A)/(2)cot.(B)/(2)=sqrt((s(s-a))/((s-b)(s-c))xx(s(s-b))/((s-c)(s-a)))=c`
`rArr (s)/(s-c)=c rArr (1)/(s-c)=(c )/(s)`
Similarly `(1)/(s-a)=(a)/(s),(1)/(s-b)=(b)/(s)`
`rArr (1)/(s-a)+(1)/(s-b)+(1)/(s-c)=(a+b+c)/(s)=(2s)/(s)=2`
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